Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using ld = long double;
const int mod = 1e9 + 7;
const ll inf = 1e9;
const ll INF = 2e18;
const int N = 1e5 + 66;
const int M = 5005;
vector<int> g[N];
const int LOG = 20;
int tin[N], tout[N], timer, up[N][LOG];
vector<int> pl[N][LOG];
void dfs(int v, int p = 1) {
  tin[v] = timer++;
  up[v][0] = p;
  for (int i = 1; i < LOG; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    pl[v][i].resize(pl[v][i - 1].size() + pl[up[v][i - 1]][i - 1].size());
    merge(pl[v][i - 1].begin(), pl[v][i - 1].end(),
          pl[up[v][i - 1]][i - 1].begin(), pl[up[v][i - 1]][i - 1].end(),
          pl[v][i].begin());
    while (pl[v][i].size() > 10) pl[v][i].pop_back();
    assert(pl[v][i].size() == 10);
  }
  for (int to : g[v])
    if (to != p) {
      dfs(to, v);
    }
  tout[v] = timer++;
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = LOG - 1; i >= 0; --i) {
    if (!upper(up[a][i], b)) a = up[a][i];
  }
  return up[a][0];
}
vector<int> get(int a, int b) {
  vector<int> vec(10, inf);
  for (int i = LOG - 1; i >= 0; --i) {
    if (!upper(up[a][i], b)) {
      vector<int> now;
      now.resize(vec.size() + pl[a][i].size());
      merge(vec.begin(), vec.end(), pl[a][i].begin(), pl[a][i].end(),
            now.begin());
      while (now.size() > 10) now.pop_back();
      assert(now.size() == 10);
      vec = now;
      a = up[a][i];
    }
  }
  {
    vector<int> now;
    now.resize(vec.size() + pl[a][0].size());
    merge(vec.begin(), vec.end(), pl[a][0].begin(), pl[a][0].end(),
          now.begin());
    while (now.size() > 10) now.pop_back();
    assert(now.size() == 10);
    vec = now;
  }
  {
    if (a != b) {
      a = up[a][0];
      vector<int> now;
      now.resize(vec.size() + pl[a][0].size());
      merge(vec.begin(), vec.end(), pl[a][0].begin(), pl[a][0].end(),
            now.begin());
      while (now.size() > 10) now.pop_back();
      assert(now.size() == 10);
      vec = now;
    }
  }
  assert(vec.size() == 10);
  return vec;
}
void solve() {
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1, a, b; i < n; ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int c, i = 1; i <= m; ++i) {
    cin >> c;
    pl[c][0].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    sort(pl[i][0].begin(), pl[i][0].end());
    while (pl[i][0].size() < 10) pl[i][0].push_back(inf);
    while (pl[i][0].size() > 10) pl[i][0].pop_back();
    assert(pl[i][0].size() == 10);
  }
  dfs(1);
  for (int a, b, l; q--;) {
    cin >> a >> b >> l;
    int c = lca(a, b);
    vector<int> vec;
    vector<int> L = get(a, c);
    vector<int> R = get(b, c);
    vec.resize(L.size() + R.size());
    merge(L.begin(), L.end(), R.begin(), R.end(), vec.begin());
    vector<int> ans;
    for (int i = 0; l && i < vec.size(); ++i) {
      if (vec[i] == inf) break;
      if (i > 0 && vec[i] == vec[i - 1]) continue;
      l--;
      ans.push_back(vec[i]);
    }
    cout << ans.size() << " ";
    for (int i : ans) cout << i << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int tt = 1;
  while (tt--) {
    solve();
    cout << "\n";
  }
}